This 'Simple' Multiplication Game Will Help You Rediscover The Joy Of Math

Like math? Like puzzles? Then meet Bojagi, a game created by David Radcliffe (@daveinstpaul).

It looks simple enough. A grid has a few numbers in it. The goal is to draw a rectangle around each number as follows: (1) the area of each rectangle is the number it contains; (2) the rectangles do not overlap; and (3) the rectangles fill the entire grid. Here is the puzzle that appears when you click on the link above.

You should try it out for yourself, but I've worked out the solution to this one and it appears at the end of this post.

Here's why I like this game. First of all, it's easy to learn and understand. In fact, a child in, say, third or fourth grade can make progress on any given puzzle. And it's a really good game for children that age. To be good at it, you must know your multiplication tables, but you also must be able to realize any given number as a product of two others. What's more, you may need to know many different ways of factoring an integer. For example, in the puzzle above you must deal with 30, which may be written as 30 x 1, 15 x 2, 10 x 3, or 6 x 5. Since this is only a 10 x 10 grid the first two are out, but either of the last two is at least possible. So this is a fun way to have your kid review factorizations of integers, a skill that is helpful for fast computations.

Another thing I like is the visual nature of the puzzle. Many people claim not to like mathematics, or that they were "terrible" at math in school. But everyone likes numbers and arithmetic when they're young, and this game helps remind you of that. Multiplication is embedded in geometry: the product of two numbers is the area of a rectangle. This visualization also helps you see that the order of multiplication doesn't matter; indeed, p x q = q x p because the rectangles whose areas they represent are the same up to a rotation by 90 degrees. But Bojagi also adds in a jigsaw puzzle aspect in which the various rectangles have to fit together to fill out the grid. So, you might have to try different factorizations of the integers (that is, make different rectangles around them) to make all the rectangles fit.

The navigation at the top right of the site includes two important links: List and Create. The first of these is a list of user-created puzzles; the second is an interface which allows you to create your own. The latter feature is really nice since it allows you to interact with the game as more than just a player. Here's one I created just now; it took about two minutes.

This simple game actually contains some fairly sophisticated mathematics, though. In particular, it involves dealing with partitions of integers. Given a positive integer n, a partition of n is a way of writing n as a sum of positive integers. For example, 4 may be written as 4, 3+1, 2+2, 2+1+1, or 1+1+1+1+1. Given n, denote by p(n) the number of partitions of n. So p(4) = 5, p(10) = 42, and so on. This enters the game as follows. The total grid has an area N and the puzzle decomposes the grid into subrectangles. This is precisely a partition of N. The first example above is a 10 x 10 grid so that N = 100. Now, p(100) = 190,569,292, so in principle there are a lot of different 10 x 10 puzzles. Not all of them are possible, however. For example, it is impossible to associate a puzzle to the partition 100 = 99 + 1 on a 10 x 10 grid since a rectangle of area 99 would have to have a side of length at least 11. Even if you allow a larger grid of area 100 it's still hopeless (that should be obvious if you think about it for a minute). The same is true for 100 = 98 + 2 and for lots of other partitions.

The partition function is very important in number theory, but there is no closed formula for evaluating p(n) for any particular n. There is a generating function--an infinite series whose coefficients are the values of the partition function--but that requires a lot of calculation to obtain a particular coefficient. There are some recurrence formulas due to Leonhard Euler, and there are asymptotic approximation formulas that allow one to estimate p(n) for large values of n.

But that's way more than you need to know to enjoy Bojagi. And, as promised, here is the solution to the first puzzle.